484 Mr. A. M c Aulay on Quaternions as a 



i. e. II is what is called the magnetic potential, A is the part 

 of the vector potential due to currents, and yA the part of 

 the magnetic force due to the same cause. The statements 

 that the parts of H and A due to magnetism are y (TfSly uds 

 and JTjv Iywcfc respectively are taken directly from the third 

 part of Maxwell's ' Electricity and Magnetism/ It remains 

 to prove that the parts of H and A due to C are yfff wCds and 

 fff w.Cds respectively. Since the * part of H due to magnetism 

 is vffJSlvwcfej the part of H contributed by a shell of 

 strength c is 



cyflWvyuds = — cjj SUvy. V u ds = c$YTJvy. yuds\_\* y 2 w=0] 



= —cffivYUvyu = cyjj YTJvyuds = CV fap [equation (1) above]. 



Hence the part of H contributed by C is vJJjwCds. This 

 proves equation (15). 



Since the part contributed to A by I is jJjYlywds, the 

 part contributed by the shell is cjj YTJvyuds=:c^udp. Hence 

 the part contributed by C is jjjwGds. This proves equation 

 (14). 



It remains to see how far Maxwell's theory agrees with 

 equations (12) to (15). Since Sv(Vv A ) = 0, we have from 

 equation (9) SyH = — 47rSyI. Hence from equation (7) 

 V H = 47r(C-S V I). 



Again, from equation (10), [SU*>H] a+6 = — 47r[SUvI] a+6 . 

 Hence from equation (8), 



[UvH] a+6 =-47r[SUvI] a+J . 

 Putting then in equation (6) <7 = H, 



H = vj^SUv Ids + VJ5JH(C-Svl)<fc 

 = VJJj(wCH-Slvw)^9 

 by equation (2). This is equation (15). 



Again, substituting A for q, and utilizing equations (9), (11), 

 4ttA = V jjjw (H + 4irl>fc - vftuYTJvrfWds 



+ V( -jJ^SUvA ds + jjJuS v Ads) . 



* In what follows I deliberately give in their quaternion form certain 

 results that might be quoted from Maxwell's ' Electricity and Magnetism,' 

 in order to show that the problem when thus worked out at full is by 

 no means a long one when treated by means of quaternions. 



